Problem: If $x$ is a real number and $\lceil x \rceil = 11,$ how many possible values are there for $\lceil x^2 \rceil$?
Answer: From $\lceil x \rceil = 11,$ we get $10 < x \le 11.$ Therefore, $100 < x \le 121,$ so the possible values of $x$ are $101, 102, \dots, 121.$ Therefore, the number of possible values of $x$ is $121 - 101 + 1 = \boxed{21}.$